如何使用接口作为“输出”参数？

Question

public interface IAnimal
{
}

public interface IDog : IAnimal
{
}

public class Dog : IDog
{
    public bool has_two_legs = false;
}

public static class test
{
    public static void QueryAnimalProperties(out IAnimal animal_details)
    {
        //some sql queries
        animal_details.has_two_legs = true;
    }

    public static void Test()
    {
        Dog my_dog;
        test.QueryAnimalProperties(out my_dog);
    }
}

When I try to call the function passing and instance of the dog class with the " out " keyword I am receiving an error:

"The best overload for method ... has some invalid arguments"

How am I able to pass a class which implements an interface to my database function to be filled with data?

UPDATE:

test.QueryAnimalProperties(out (IAnimal)my_dog);

Trying to type cast the input also gives an error:

A ref or out argument must be an assignable variable

Answer 1

QueryAnimalProperties could return an object that is IAnimal , but not IDog (eg ICat ). Such an object would not be assignable to an IDog variable. Therefore, this is forbidden.

Answer 2

You don't need an out parameter.

But if you want to use it, then use a Generic Method with constraints.

public interface IAnimal
{
    string Name { get; set; }
}

public interface IDog : IAnimal
{

}

public void QueryAnimalProperties<T>(out T animal)
where T : IAnimal, new()
{
    animal = new T();
    animal.Name = "Fred";    
}    

public class Dog : IDog
{
    public string Name { get; set; }
}

void Main()
{
    Dog dog;
    QueryAnimalProperties(out dog);
    Console.WriteLine(dog.Name);
}

Note that if you remove the out parameter without modifying the rest of your code, then your application basically depends on what is known as side-effect, which is something you want to avoid in this situation.

http://codebetter.com/matthewpodwysocki/2008/04/30/side-effecting-functions-are-code-smells/

Answer 3

There's no reason to use the out parameter here.

Remove it from your method, and just update the properties of the class you pass to it (that are available in the IAnimal interface), if that's what you're intending to do.

public void QueryAnimalProperties(IAnimal animal_details)
{
   //some sql queries

   animal_details.SomeAvailableProperty = "SomeValue";
}

Because the class is already a reference type, when execution returns to whatever method called QueryAnimalProperties , your instance of Dog will retain the values you set in it.

Answer 4

You don't pass an instance of an object as an input value for an out parameter, the value will be discarded anyway.

Here's a simple counter-example that demonstrates why it doesn't work:

public void QueryAnimalProperties(out IAnimal animal_details)
{
   animal_details = new Hamster();
}

//

Dog dog;
QueryAnimalProperties(out dog);

In this case, the implementation of QueryAnimalProperties is valid because Hamster : IAnimal and animal_details can accept an object of type Hamster as it implements the interface.

But the caller of QueryAnimalProperties cannot expect animal_details to ultimately resolve as Dog , as my example shows, it might set it to Hamster .

As an aside, I don't think you're using out correctly: out means that the reference itself is changed, as though it were a pointer-to value in C and C++ (or as this is a heap object, it would be pointer-to-pointer value).

Consider out params as equivalent to the return-type of the function, change your function from this:

public void QueryAnimalProperties(out IAnimal animal_details)

to this:

public IAnimal QueryAnimalProperties()

Dog dog = QueryAnimalProperties(); // invalid, QueryAnimalProperties is not guaranteed to return Dog.

and it works the same way.

Answer 5

This fails becuase Dog is a concrete class

   public static void Test()
    {
        Dog my_dog; // <-- here
        test.QueryAnimalProperties(out my_dog);
    }

The out variable animal_details in

public static void QueryAnimalProperties(out IAnimal animal_details)

just promises an IAnimal not necessarily a Dog

To fix this edit the type from Dog to IAnimal :

   public static void Test()
    {
        IAnimal my_dog; // <-- here
        test.QueryAnimalProperties(out my_dog);
    }